= Solution
The circular <eigenmode> <orthogonal projection> matrices obey $P_\pm^2=P_\pm$, $P_+P_-=0$ and $P_++P_-=I$. Define the uniform-medium generator
$$
K=\gamma^+P_++\gamma^-P_-
=\boxed{\begin{pmatrix}\gamma&i\delta\\-i\delta&\gamma\end{pmatrix}.}
$$
Expanding its <matrix exponential> by its <power series> and using these projection identities gives
$$
\boxed{e^{-Kz}=e^{-\gamma^+z}P_++e^{-\gamma^-z}P_-=\mathcal J(z).}
$$
Equivalently, with $Q=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$, one has $Q^2=I$, $K=\gamma I+\delta Q$ and $e^{-Kz}=e^{-\gamma z}[I\cosh(\delta z)-Q\sinh(\delta z)]$. This is exactly the corrected matrix of part (d). The entries of $K$ have units of inverse length, and the map satisfies $d\mathcal J/dz=-K\mathcal J$, $\mathcal J(0)=I$ and the homogeneous propagation law $\mathcal J(z_1+z_2)=\mathcal J(z_1)\mathcal J(z_2)$.
The extra $\operatorname{sech}^2$ factor in the printed version violates this last law for nonzero $\delta$, so it cannot be generated by one constant $K$ with the stipulated <eigenvalues>. The constant-$K$ description assumes uniform coefficients and fixed circular <eigenmodes>.
Back to article page
